My Latin teacher is a retired nuclear physicist. He is a very smart guy. Sometimes, he shows us things called ABL (anything but latin) like math or logic. Its a fun class. Anyway, there is one geometry problem he gave us that has been bugging me as well as Apotheosis.

Here is my solution - but note that I am using ^ to indicate angles (as opposed to triangles).

To prove that ABC is isoceles, we need to prove at least one of the following:
AC = BC
^CAB = ^CBA

So -
Let us add to the diagram segment DF.
This gives us quadrilateral ADFB.
Since DB and AF are congruent diagonals of ADFB which bisect the base angles DAB and FBA,
ADFB must be an isoceles trapezoid.

The lower base angles of any isoceles trapezoid are equal, therefore
^DAB = ^FBA
Since
^DAB = ^CAB and ^FBA = ^CBA
it goes to show that
^CAB = ^CBA
Thus, ABC must be isoceles.

But how do you know that DF is parallel to the base segment. If it were, then angle CAB must be congruent to angle CBA. You cannot assume this to prove it. In other words, you are assming that the two base angles are congruent. This is what we are trying to prove!

To show you what I mean,

Since DB and AF are congruent diagonals of ADFB which bisect the base angles DAB and FBA, ADFB must be an isoceles trapezoid.

You are assuming that the angles bisected are congruent. Their bisectors may be equal in length, but the angles may not be the same. Therefore the top of the quadrilateral is not paralell to AB.

An isoceles trapezoid has the following properties:
it is a quadrilateral
it has congruent diagonals
its diagonals bisect its internal angles
the lower base angles are congruent
the upper base angles are congruent
the pair of non-parallel edges are congruent
the pair of parallel edges are not congruent

Thus, if ADFB can be proven to be an isoceles trapezoid, ADFB has all these properties. That would show that DF is parallel to the base segment, that the base angles are equal, and that my solution holds.

From the information given, we know that ADFB has congruent diagonals which bisect the lower base angles. I have taken this to prove that ADFB is an isoceles trapezoid; you are questioning whether that is proof enough.
In other words, you are asking:
Can't there exist a quad which has congruent diagonals that bisect the internal angles, but which is NOT an isoceles quadrilateral?I've tried looking that up, but so far haven't managed to find anything about it. From testing it so far, it seems the answer to the question is no - but if anyone can show otherwise, I'm open to persuasion.

...
So - I will accept, for the moment, that may proof might not hold (though it isn't ready to be tossed out the window yet!).

We know that the two diagonals are congruent. But how do you prove that the angles that they bisect are congruent? Its the whole point of the problem. Like I said in my 1st post, we cannot assume anything.

I tried almost the exact proof and showed it to my teacher. That is what he pointed out.

P.S. Just to let you guys know, I'm trying to figure this out just so my teacher will give me a buck. LOL! Its a challenge though, huh?

Raya wrote:From the information given, we know that ADFB has congruent diagonals which bisect the lower base angles. I have taken this to prove that ADFB is an isoceles trapezoid

This is only the case if the bisected angles are congruent. This is only the case if the triangle ABC is isosceles, which is what is to be proven. Threewood is correct; you are assuming what is to be proven, without realizing it.

I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!

Look at triangles ACF and BCD. If we can prove that these triangles are congruent, then we prove that triangle ABC is isosceles! In order to prove that a triangle is congruent to another, we must prove a combination of either 3 angles or sides. For example we can have ASA or SAS. Let's try to do this for these 2 triangles.

First, both triangles share a common angle. This angle is angle ACB. Angle ACB is congruent to angle ACB by the reflexive property. We know that segments AF and BD are congruent. This information is given. In order to prove that the 2 triangles are congruent, we must prove one of the following:

CD is congruent to CF
AC is congruent to CB
Angle CAF is congruent to angle CBD
Angle CDB is congruent to angle CFA

If anyone of these can be proven, then all the others can be proven true by CPCTC (Congruent Parts of Congruent Triangles are Congruent)

I'm not sure if the Law of Sines works because I haven't figured it out for myself. It's tougher than one thinks...

I do not think that triangles with two common sides must themselves be congruent. You also need an angle or another side. I can have two triangles. 4 of the 6 sides are congruent. Okay its hopeless trying to explain in words. Click here to see another diagram...

Jung He Fah Toy wrote:If either of the triangles Lex and threewood were pointing out can be proven congruent, then the problem can be solved!

I tried several ways, to no avail. I suspect the constraints are insufficient to force the triangle to be isosceles. If this is the case, then it should be possible to find a counter-example that meets all the constraints and yet is not isosceles. This would at least prove that it's a trick question.

I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!

Actually, wrote that question on my drill today before I handed it in. I said: "Can you use the law of sines to prove this triangle to be isosceles? Can you use a geometric proof to prove it is isosceles? Is it even possible?"

I hope he gets back to me tomorrow, if not we will aks him personally.

First, let's start with a slightly different problem. Let's assume that DB and AF are "altitudes" of the triangle. That is, DB meets AC at a 90 degree angle, and AF meets BC at a 90 degree angle. It can be shown that if DB and AF are congruent, then the triangle must be isosceles. (I don't know the proof for this; I found it in a geometry book, and the proof wasn't included.)

OK, if you have that isosceles triangle with congruent altitudes DB and AF, then by pivoting DB and AF by the same number of degrees (and extending the line segments as necessary), you should be able to bisect the angles ABC and BAC. DB and AF will remain congruent. This proves that if ABC is an isosceles triangle with two line segments bisecting the congruent angles, then the bisectors will be congruent as well.

Now, can the converse be proven by converting the bisectors to altitudes? If so, the problem is solved. I'll leave this as an exercise for the reader.

I, Lex Llama, super genius, will one day rule this planet! And then you'll rue the day you messed with me, you damned dirty apes!

I almost got it. Sin A/AF = Sin B/DB. AF is con. to DB. So SinA/AF = SinB/AF. THerefore angles CAB and CBA are congruent. Its not correct though trust me.

I found online that you have to use the apex, which in this case is angle C, and the law of sines to prove this. Look at triangles CDB and CFA. Try to do something with this so prove that these triangles are congruent.

Kyle, I have something that will bust this mystery wide open (I hope). I showed this problem to my parents, and my mom went and got her old high school math book. Sure enough, the problem that Mr. Sytek gave us is in there! However, in the book, it is given that ?ABC is isosceles and you must prove AF and DB to be congruent. Now, the problem we are trying to solve is the exact opposite of that! We are trying to prove ?ABC to be isosceles given that AF and DB are congruent. The way I see it, if we can somehow "reverse engineer" the proof in the book, then we can solve this problem once and for all. I will bring the book into class tomorrow so you can see it. Hopefully we will be able to figure it out! Hehehehe....

P.S. - Mr. Sytek simply wrote on my paper: It is possible, but it is very difficult to prove it.

Yes but we already know this. We know that the triangle can exist no other way.

We are trying to prove that it is an isosceles triangle not by proving that if it were any other way it would not be, but by using the info provided. In other words, we need a

this is this
threrfore this is this
therefore this equals this
therefore this is this

and so on

The challenge is proving it this way. It is possible with the law of sines. Our Latin teacher told us (who by the way is a retired nuclear physicist) that is is possible using the law of sines. It is harder than one thinks...